if 51.0L of a gas is collected at 29C and 305kPa, what is its volume at STP?

To find the volume of the gas at STP (Standard Temperature and Pressure), we need to use the ideal gas law equation, which is:

PV = nRT

Where:
P = Pressure of the gas
V = Volume of the gas
n = Number of moles of gas
R = Ideal gas constant
T = Temperature of the gas in Kelvin

Given information:
Initial Volume (V1) = 51.0 L
Initial Temperature (T1) = 29°C = 29 + 273 = 302 K
Initial Pressure (P1) = 305 kPa
Standard Temperature (T2) = 0°C = 0 + 273 = 273 K
Standard Pressure (P2) = 101.3 kPa (STP)

Now, we need to calculate the number of moles of the gas using the ideal gas law equation and the given values:
PV = nRT

n = (PV) / (RT)

n = (305 kPa * 51.0 L) / ((0.0821 L·atm/(mol·K)) * 302 K)

n ≈ 6.41 mol

Next, we can use the number of moles and the new pressure and temperature at STP to find the new volume (V2):
PV = nRT

(V2)(101.3 kPa) = (6.41 mol) * (0.0821 L·atm/(mol·K)) * (273 K)

V2 = [(6.41 mol) * (0.0821 L·atm/(mol·K)) * (273 K)] / (101.3 kPa)

V2 ≈ 11.1 L

Therefore, the volume of the gas at STP is approximately 11.1 L.